I'm performing speech intelligibility tests in a cross-over design study. In effect, a paired t-test, or repeated-measures ANOVA analysis is the way to go here, because every subject serves as their own control. Basically, we test every subject in a control condition (C) and in a condition where we have tempered with the acoustic input to their hearing implant (T1, T2...). The outcome measure is the speech-reception threshold (SRT). The SRT is basically the signal-to-noise ratio where 50% of the words has been identified correctly.

So basically we can do paired t-tests or RM ANOVAs with the absolute SRTs as input.

We can also pre-determine the difference values. For both the RM ANOVA and the t-tests the inputs would be C-C (always yielding 0), T1-C, T2-C etc.

This made me wonder:

Do paired t-tests and RM-ANOVA yield the same result as their unpaired counterparts with the differences as input? (SRTcontrol - SRTtest).

  • 1
    $\begingroup$ (+1) A paired t-test is exactly equivalent to the one-sample t-test on the differences. But for the RM-ANOVA with more than 2 groups it's not so clear what "unpaired counterpart" you mean. $\endgroup$ – amoeba Jun 26 '17 at 10:23
  • $\begingroup$ @amoeba thank you! I tried to clarify the question. Your comment does make me wonder whether including the control-minus-control condition makes sense, as its mean value and the variance will always equal zero. $\endgroup$ – AliceD Jun 26 '17 at 10:31
  • 1
    $\begingroup$ Exactly. After you subtracted control condition, you basically don't have it anymore (that's why a paired two-sample t-test is equivalent to a one-sample test on the differences: the differences are compared with 0, not with some other data). So if you had C, T1, and T2 in the original RM-ANOVA, you will now have T1-C and T2-C and would need to devise some procedure to test the null that they are both zero. I doubt that it's possible to do such that the result is fully equivalent to the original RM-ANOVA. But even if it is somehow possible, it's definitely not some standard "counterpart". $\endgroup$ – amoeba Jun 26 '17 at 11:06
  • $\begingroup$ @amoeba - I'm unsure about the conventions here on this site (what's an answer, what's not), but your comment is basically the answer to my question, regardless site's conventions. If you like to answer it, please go ahead and I can mark it as accepted. The current answer is nice, but not an answer to my specific inquiry. $\endgroup$ – AliceD Jun 27 '17 at 7:56

The answer is no. The paired tests are identical to one-sample tests as amoeba pointed out in his comment, but not identical to the unpaired test.

The unpaired t-test assumes two independent samples. So the variance of the mean difference is the sum of the variances the means have in each sample:

$$Var(\bar{X}_1 - \bar{X}_2) = Var(\bar{X}_1) + Var(\bar{X}_2)$$

Conversely, paired observations are not (necessarily) independent, so the variance of their mean difference is

$$Var(\bar{X}_1 - \bar{X}_2) = Var(\bar{X}_1) + Var(\bar{X}_2) - 2Cov(\bar{X}_1, \bar{X}_2).$$

One just doesn't see this more complicated variance since it is in fact easier by the pairing: Estimate the variance on the left side just by the per-subject differences. And this makes it identical to the one-sample test.

This is similar for the more general case of repeated measures or multiple groups.

  • 1
    $\begingroup$ Isn't it though, if you assume the null hypothesis is 0 and it's a two-group case? Your standard error is calculated on the differences. $\endgroup$ – HEITZ Jun 26 '17 at 23:56
  • $\begingroup$ Yes. I think if I show that repeated measures and multiple groups are different for the special case of two repeated measures (then called paired data), it can't be similar in the more general case of 2 or more repeated measures. Mathematician's rule: Choose simple counterexamples. ;) $\endgroup$ – Horst Grünbusch Jun 28 '17 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.